User:Polygon/Page for testing: Difference between revisions

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{{TM|1RB3RB5RA1LB5LA2LB_2LA2RA4RB1RZ3LB2LA|halt}} is the current [[BB(2,6)]] [[champion]]. It was discovered on the 19th of May 2023 by Pavel Kropitz. It halts with a score > <math>10 \uparrow\uparrow 10 \uparrow\uparrow 10^{10^{115}}</math>.
==Analysis by Shawn Ligocki==
https://www.sligocki.com/2023/05/20/bb-2-6-p3.html
<pre>
Analysis
Level 1
These rules can all be verified by direct simulation:
00 <A 212 22^n 55 → <A 212 22^n+2


00 <A 212 22^n 2 55 → <A 212 55^n+2 2
0^5 <A 212 22^n 52 5555 → <A 212 55 2 55^n+3 52
00 <A 212 22^n 2 52 5 → <A 212 55^n+2 52
Level 2
Repeating the first rule above we get:
0^∞ <A 212 22^n 55^k → 0^∞ 212 22^n+2k
which let's us prove Rule 2:
0^∞ <A 212 22^n 2 55 → 0^∞ <A 212 55^n+2 2
                    → 0^∞ <A 212 22^2n+4 2
Level 3
Repeating Rule 2 we get:
0^∞ <A 212 22^n 2 55^k → 0^∞ <A 212 22^(n+4)*((2^k)-4) 2
which let's us prove Rule 3:
0^∞ <A 212 22^n 52 5^5 → 0^∞ <A 212 55 2 55^n+3 52 5
                      → 0^∞ <A 212 22^2 2 55^n+3 52 5
                      → 0^∞ <A 212 22^(6*2^(n+3)-2) 52 5
                      → 0^∞ <A 212 55^(6*2^(n+3)-2) 52
                      → 0^∞ <A 212 22^(6*2^(n+4)-4) 52
Level 4
Let
f(n) = 6*2^(n+4)-4
Repeating Rule 3 we get the Tetration Rule:
0^∞ <A 212 22^n 52 5^5k → 0^∞ <A 212 22^f^k(n) 52
This rule will be the main contributor to the score since f^k(n) > 2^^k. In fact, this rule will apply 3 times, which is how we end up with 3 tetrations in the final score (>10^^10^^10^^3).
</pre>
∞∞∞∞
→→→→

Latest revision as of 17:31, 26 September 2025

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